f(x) = 2x^2 - 4x + 3
First we need to determine f'(x)
f'(x) = 4x - 4
Now we need to find the critical values for f(x) which is f'(x)'s zero.
4x - 4 = 0
==> 4x = 4
==> x= 1
Then f'(x) has a minimmum values...
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f(x) = 2x^2 - 4x + 3
First we need to determine f'(x)
f'(x) = 4x - 4
Now we need to find the critical values for f(x) which is f'(x)'s zero.
4x - 4 = 0
==> 4x = 4
==> x= 1
Then f'(x) has a minimmum values at x= 1
==> f(1) = 2(1) - 4(1) + 3
= 2 - 4 + 3
= 1
Then f has a minimmum value at (1, 1)
Find the minimum by taking the derivitive of the function and setting it to zero.
Use the power rule which states: d/dx [x^n] = n*x^(n-1)
f(x) = 2x^2 - 4x^1 + 3
f'(x) = 4x -4
Now set to zero
4x - 4 = 0
x = 1
minimum is at f(1) = 2 - 4 + 3 = 1 , or the point (1,1)